Problem: Six congruent circles form a ring with each circle externally tangent to the two circles adjacent to it.  All six circles are internally tangent to a circle $\cal C$ with radius 30.  Let $K$ be the area of the region inside $\cal C$ and outside all of the six circles in the ring.  Find $\lfloor K\rfloor$. (The notation $\lfloor K\rfloor$ denotes the greatest integer that is less than or equal to $K$.)
Solution: Let $r$ be the radius of each of the six congruent circles, and let $A$ and $B$ be the centers of two adjacent circles. Join the centers of adjacent circles to form a regular hexagon with side $2r$.  Let $O$ be the center of $\cal C$.  Draw the radii of $\cal C$ that contain $A$ and $B$. Triangle $ABO$ is equilateral, so $OA=OB=2r$.  Because each of the two radii contains the point where the smaller circle is tangent to $\cal
C$, the radius of $\cal C$ is $3r$, and $K=\pi\left((3r)^2-6r^2\right)=3\pi r^2$. The radius of $\cal C$ is 30, so $r=10$, $K=300\pi$, and $\lfloor K\rfloor=\boxed{942}$.